$ \int_{-1}^0 \int_{-2 - 2y}^{1 + y} dx \, dy + \int_0^1 \int_{-2 + 2y}^{1 - y} dx \, dy$ Switch the bounds of the double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^2 \int_{-1 - x}^{1 + x} dx \, dy + \int_{-1}^0 \int_{-1 + x/2}^{1 - x/2} dx \, dy$ (Choice B) B $ \int_{-2}^0 \int_{-1 - x}^{1 + x} dx \, dy + \int_0^1 \int_{-1 + x/2}^{1 - x/2} dx \, dy$ (Choice C) C $ \int_{-2}^0 \int_{-1 - x/2}^{1 + x/2} dx \, dy + \int_0^1 \int_{-1 + x}^{1 - x} dx \, dy$ (Choice D) D $ \int_0^2 \int_{-1 - x/2}^{1 + x/2} dx \, dy + \int_{-1}^0 \int_{-1 + x}^{1 - x} dx \, dy$
The first step whenever we want to switch bounds is to sketch the region of integration that we're given. We are given two regions in this case. The first has $-2 - 2y < x < 1 + y$ and $-1 < y < 0$. The second has $-2 + 2y < x < 1 - y$ and $0 < y < 1$. Therefore: ${1}$ ${2}$ ${\llap{-}2}$ ${1}$ ${2}$ ${\llap{-}2}$ $y$ $x$ Because we're switching bounds to $dy \, dx$, we need to start with numeric bounds for $x$. We'll need two regions, so we'll split our calculations into two parts. For the first region, we see that $-2 < x < 0$. Now we can define bounds for $y$ as $-1 - \dfrac{x}{2} < y < 1 + \dfrac{x}{2}$. For the second region, we see that $0 < x < 1$. We can define bounds for $y$ as $-1 + x < y < 1 - x$. We want to pay attention especially to how these $y$ bounds works at the edges of the $x$ bounds. For example, at $x = -2$, the first $y$ bound makes $y = 0$ as expected. In conclusion, the double integral after switching bounds is: $ \int_{-2}^0 \int_{-1 - x/2}^{1 + x/2} dx \, dy + \int_0^1 \int_{-1 + x}^{1 - x} dx \, dy$